Question

(a) Define the following terms:
(i) ideal solution (ii) Azeotrope (iii) Osmotic pressure
(b) A solution of glucose ${{C}_{6}}{{H}_{12}}{{O}_{6}}$ in water is labelled as $10$ by weight. What would be the molality of the solution?
( molar mass of glucose = $180g\text{ }mol{{e}^{-1}}$ )

Answer 1

First of all, to answer this you have to recall all the definitions of the solutions chapter and then you can easily define the given terms and the molality of the solution can be found by applying the formula as: $Molality=\dfrac{No.\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{mass\text{ }of\text{ }the\text{ }solvent\ in\text{ }grams}\times 1000$. Now answer the statement.

Complete Solution:
Let’s discuss the given statements one by one as:
(a) (i) By the ideal solutions, we mean those which obeys the Raoult’s law over the entire range of the concentration.
Now what is Raoult’s law? It may be defined as the vapor pressure of the solution containing non-volatile solute is directly proportional to the mole fraction of the solvent. i.e.
${{p}_{A}}=p_{A}^{\circ }{{x}_{A}}$ and ${{p}_{B}}=p_{B}^{\circ }{{x}_{B}}$.
For a solution to be ideal it should satisfy the following three conditions:
1. It should obey Raoult’s law.
2. There should be no change in enthalpy on mixing the two components i.e. ${{\Delta }_{mixing}}H=0$.
3. There should be no change in volume on mixing the two components i.e. ${{\Delta }_{mixing}}V=0$.

(ii) The solutions i.e. the liquid mixture which boil at constant temperature and can distill unchanged in composition are called azeotropes or the azeotropic mixtures. They are of two types;
1. Minimum boiling azeotropes:- Solutions which show positive deviations from ideal behavior form minimum boiling azeotropes i.e. ${{p}_{A}}>p_{A}^{\circ }{{x}_{A}}$ and ${{p}_{B}}>p_{B}^{\circ }{{x}_{B}}$. ${{\Delta }_{mixing}}H = +ve$ and ${{\Delta }_{mixing}}V=+ve$

2. Maximum boiling azeotropes:- Solutions which show negative deviations from ideal behavior form maximum boiling azeotropes i.e. ${{p}_{A}}$<$p_{A}^{\circ }{{x}_{A}} and {{p}_{B}}$<$p_{B}^{\circ }{{x}_{B}}$. ${{\Delta }_{mixing}}H = -ve$ and ${{\Delta }_{mixing}}V=-ve$

(iii) The equilibrium hydrostatic pressure on the solution due to the osmosis of the pure solvent into it is a measure of osmotic pressure.
We may also define osmotic pressure as the excess pressure which must be applied to the solution to prevent the passage of solvent into it through a semipermeable membrane. It is generally denoted by the symbol $\pi$.

(b) $10$ by weight means 10 gram of the glucose is present in the 100 gram of the solution.
As we know that:
$Molality =\dfrac{No.\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{mass\text{ }of\text{ }the\text{ }solvent\ in\text{ }grams}\times 1000$
Now, we know that mass of the glucose = $180g\text{ }mol{{e}^{-1}}$(given)
Given mass of the glucose = 10 gram
So, $\begin{aligned} & Moles\text{ }of\text{ }glucose=\dfrac{given\text{ }mass}{molecular\text{ }mass} \\\ & \text{ =}\dfrac{10}{180} \\\ & \text{ = 0}\text{.0556} \\\ \end{aligned}$
Mass of the solvent = $100 - 10 = 90\text{ gram}$
Then,
$\begin{aligned} & Molality = \dfrac{0.0556}{90}\times 1000 \\\ & \text{ = 0}\text{.618 m} \\\ \end{aligned}$
Therefore, the molality of the solution in water = 0.618m

Note: Molality of the solution is always independent of the temperature and does not change with the change in the temperature because the mass of the solvent is independent of temperature and hence, the molality.